Question

given ((x - 7)^2 = 36), select the values of (x).
(\boldsymbol{checkmark}) (x = 13)
(\boldsymbol{checkmark}) (x = 1)
(\boldsymbol{\times}) (x = -29)
(\boldsymbol{\times}) (x = 42)
complete

given ((x - 1)^2 = 50), select the values of (x).
(square) (x = -49)
(square) (x = 51)
(square) (x = 1 + 5sqrt{2})
(square) (x = 1 - 5sqrt{2})
done (\boldsymbol{checkmark})

Explanation:

Step1: Solve first equation: take square roots

Take square root of both sides of $(x-7)^2=36$:
$$x-7=\pm\sqrt{36}=\pm6$$

Step2: Solve for x (two cases)

Case 1: $x-7=6$
$x=6+7=13$
Case 2: $x-7=-6$
$x=-6+7=1$

Step3: Solve second equation: take square roots

Take square root of both sides of $(x-1)^2=50$:
$$x-1=\pm\sqrt{50}=\pm5\sqrt{2}$$

Step4: Solve for x (two cases)

Case 1: $x-1=5\sqrt{2}$
$x=1+5\sqrt{2}$
Case 2: $x-1=-5\sqrt{2}$
$x=1-5\sqrt{2}$

Answer:

For $(x-7)^2=36$:

• $x=13$
• $x=1$

For $(x-1)^2=50$:

• $x=1+5\sqrt{2}$
• $x=1-5\sqrt{2}$